1:29 AM
@Slereah Heh, that notation is a bit... self-defining, no?

4 hours later…
5:30 AM
Indices are a bit funny above, if $A = A_{\mu} dx^{\mu}$ is a one-form, it's exterior derivative is $F = dA = \partial_{\nu} A_{\mu} dx^{\nu} \wedge dx^{\mu} = \frac{1}{2} (\partial_{\nu} A_{\mu} - \partial_{\mu} A_{\nu} ) dx^{\nu} \wedge dx^{\mu} = \partial_{[\nu} A_{\mu]} dx^{\nu} \wedge dx^{\mu} = 2 F_{\nu \mu} dx^{\nu} \wedge dx^{\mu}$
where one goes from $\partial_{\nu} A_{\mu}$ to $\frac{1}{2} (\partial_{\nu} A_{\mu} - \partial_{\mu} A_{\nu})$ because the $dx^{\nu} \wedge dx^{\mu}$ is anti-symmetric, but $\partial_{\nu} A_{\mu}$ isn't however it can be made to be with the $1/2$ thing
(Should be a $1/2$ infront of the $F_{\nu \mu}$)

6:10 AM

6:22 AM
Hello good people. in my effort to understand better the difference between the relative vector and the reduced mass and the center of mass vector with the total mass in a system of two objects i am looking for a visualization of this phenomena maybe with vectors pointng at each while the objects move and interact. anyone know something similiar ?

6:39 AM
thanks. i also find htis visualization youtube.com/…

@DanielSank You can use it with the Leibniz rule to generalize it!

7:23 AM
I think inhabiting a hill is really inconvenient unless you have a mechanical car.

7:41 AM
if those shops putting a lot of chairs and tables for everyone to use freely also put a lot of computers on those tables for everyone to use freely, it's really a helpful favor.
my MSc school is also on a hill, but there were a lot of eateries and other shops on that hill so that it was not so inconvenient to inhabit there.
but this hill I am located on only has factories and gallery, and has no one single eatery or shop.

8:03 AM
Is this really correct? en.wikipedia.org/wiki/…

8:57 AM
My first attempt at data science. I downloaded a kaggle dataset and plotted everything with python.

9:53 AM
0

Suppose I have in a constant gravitational field a shallow water tray that empties at height $h$ into a round pipe that opens underwater in another shallow tray at height $0$. My intuition suggests that, if we assume laminar flow and the trays' water levels are kept constant, and the pipe cross-s...

Anyone can shed any light on this question?

1 hour later…
11:01 AM
If the exterior derivative doesn't require a metric, is it metric independent or does a choice of connection change it?
sure it's metric independent otherwise equations like $\text dA=F$ get messed up on non-flat backgrounds
or actually any background for that matter, I guess the flat qualifier isn't needed

It is independent of the connection, yes
Although be aware that not all exterior algebra is
ie the Hodge star isn't
that's why EM still depends on the metric somewhat

how come the hodge star is metric dependent if it doesn't allow you to move between forms and blades?
in my head that's the operation that requires a metric to define the map between $V$ and $V^*$
maybe the answer is more complicated than I can understand right now

It is defined as $a \wedge \star b = \langle a, b \rangle \omega$

ahhh

You have an inner product, that's where the metric comes in

11:09 AM
that formula looks familiar

@Charlie There are connections that influence the exterior derivative, cf. the exterior covariant derivative, but the Levi-Civita connection doesn't because it's symmetric/torsion-free and hence the antisymmetry of the exterior derivative just kills off the term you might try to add

ok that makes sense ty

@Charlie The Hodge dual is a generalization of the notion of assigning a normal vector to a surface, and the notion of what "normal" means requires a metric

Oh right, you're supposed to add the whatever form
Torsion form?

are the covariant exterior derivative and regular ex. derivative distinct objects? or is the regular ex. derivative more of a special case?

11:12 AM
Well, given a connection form $A$, the covariant derivative is $\mathrm{d}_A = \mathrm{d} + A$, so yes it's a different object
You will see this object when phrasing gauge theories like electromagnetism/Yang-Mills in coordinate-free form

ok nice

It's too hot to do physics
I have a plan

Stay in a shop with air conditioner.

But then I will catch the corona

But the problem is no shop of this kind provides free computer to do physics.
I don't think coronavirus is around in the shop I am in.

11:25 AM
isn' t the hodge star a unary operator?

Yes

the definition $a \wedge \star b = \langle a, b\rangle \omega$ requires two forms though

It maps $p$-forms to $(n-p)$-forms
@Charlie Note that the star only applies to $b$ though

or rather two objects
is it linear over the exterior product?
$a\wedge \star b = \star(a\wedge b)$?
even then surely $a$ and $b$ have to be related some how otherwise it's not uniquely defined on $b$

$a \wedge b$ is a $2p$-form, it may be zero :p
They don't have to be, no
You can define the star operator by itself if you prefer
but it's less pleasant

11:29 AM
so $a$ in the definition is just an arbitrary $p-$form?

Yes
If you want the component-wise definition, it's $$(\star \alpha)_{i_{k+1}, \dots, i_n} = \frac{\sqrt{|\det [g_{ab}]|}}{k!} \alpha^{i_1, \dots, i_k}\, \,\varepsilon_{i_1, \dots, i_n}.$$

oh yikes

@Charlie no, it's a n-p form if $b$ is a p-form
and the $\omega$ is the top-dimensional volume form - the definition makes no sense if the dimensions don't add up to n

oh sure otherwise the inner product isn't defined

As you can see the hodge star depends on the metric in two ways
The metric determinant, and also the indices are raised

11:31 AM
ok that makes it clear where the metric enters
I still can't get over the fact that $a$ is arbitrary lol

Well coordinate-free definitions are like that a lot
For instance, raising indices are defined similarly
If you have a $1$-form $\omega$, the corresponding vector $X$ is defined by $$g(X, Y) = \omega(Y)$$
That's because, as you may remember, forms are basically functions
on vectors
and similarly, vectors can be defined as funtions on forms
Tensors overall are defined by how they act on vectors and forms, as a result

this might be silly, but why can't $$\omega(\cdot)=g(X,\cdot)$$ then we don't have to pick a specific $Y$

Sure, you can define it like that
But that's basically the same idea
Partial application, as it is called

oh it is
then it just becomes $$\cdot\wedge \star b=\langle \cdot,b\rangle \omega$$
I guess I can live with that

if you so wish

11:37 AM
:P

The benefit of components is that it's a bit more explicit what that function is like

so analogous to how $g_{\mu\nu}v^\nu$ doesn't include a vector field, but is defined by how it acts on one

yes

ok nice

via Einstein notation, you can immediatly see what happens by applying any vector to it

12:12 PM
when we write maxwell's equations in terms of forms, a few new symbols are used, we end up with $D$ and $H$ for instance. Is the reason for this that in ordinary vector calculus we are making the identification ACM mentioned between 2-blades and vectors and so $D$ and $E$ for instance become heuristically "the same" object?

I don't know what you mean by that
Maxwell's equations in coordinate-free form are just $\mathrm{d}F = 0$ and $\mathrm{d}{\star}F = \star J$
(modulo constants)

That $D$ and $H$ are the fields you need when formulating Maxwell's equations inside a medium, nothing to do with the coordiante-free variant

so for instance instead of $\int_{\partial V} \vec E = \int_V \rho$ we have $\int_{\partial V} D = \int_V \rho$
ah

note how it says "Maxwell's equations in matter" in bold in the text :P

12:16 PM
ah yes it does say that lol

also that's not written in terms of forms, that's just ordinary vector calculus

ah that actually explains something I saw earlier then, ty

I feel phone is not a good device to do physics.

2 hours later…
2:08 PM
even when there is no sunshine at night, I feel so hot.

2:33 PM
Is anyone feeling in a "vote to reopen" mood?
Bhavay asked this question and I edited it to clarify because it's something widely claimed in books that seems very hard to prove (or maybe the proof is easy and I've just never read it):
2

In many introductory books on electrostatics you can find the statement that the field inside a conducting shell is zero if there are no charges within the shell. For example if we place an uncharged conducting sphere in a uniform electric field we would get something like this: For simplicity I...

It got closed as homework, which seems baffling as it obviously isn't. I am curious to know the answer and will put a large bounty on it if it's reopened. So if anyone else also wants to know the answer, or knows the answer and wants a large bounty please vote to reopen.
@FakeMod you were one of those voting to close. How is this homework like?

@JohnRennie While I'm willing to believe that your edit is what the author intended to ask all along, it invalidates the existing answers talking about "penetration" because the edited version no longer asks about the electric field penetrating anything
Wouldn't it be better to just post the edited version as a new question?

Bhavay and I collaborated on the rewrite. It was what he intended but English is not his native language.
Isn't reposting a closed question contrary to at least the spirit of the site rules if no the letter?

2:49 PM
Yes, but editing a question so that its existing answers become incomprehensible is even more so

The only undeleted answer was posted after the edit (though it is largely incomprehensible)
The impenetrability of that answer is not due to any editing on my part :-)

Hmmm, you're right
also physics.stackexchange.com/q/356444/50583, I feel this question has been asked in various forms already more times than necessary - put a bounty on one of these if you don't like the detail of the existing answers?

@ACuriousMind it's not the electric field in a conductor, it's the electric field in a void in a conductor.

for the record, I'd agree this isn't homework, I just don't want to reopen it just to have it be closed as a dupe shortly afterward

Only one of the three links you posted is even close to relevant.
Anyway, why is this so hard? It seems such a simple question!
Can it really be none of the experienced physicists hereabouts knows the answer?

2:59 PM
@JohnRennie What's wrong with ZtH's answer physics.stackexchange.com/a/356470/50583 to the third one

@ACuriousMind he just states the field is zero with no proof, or even justification.
And his picture shows a void with a charge in it, which is exactly what was not asked.

@JohnRennie By the same argument we use for the solid conductor, no? The charges in a conductor are free to move - if the field wasn't zero, they would move so that it is
I don't see why the void would change the nature of the argument

The void inside the shell isn't conducting and there an be field lines within it.
It's the field in the void I'm asking about, not in the metal of the conductor.

yes, but inside the void there are no sources - the only source for the field is outside/on the boundary of the conductor

There are three sets of charges:
1. the external charge
2. the induced charge on the outer surface of the shell
3. the induced charge on the inner surface of the shell
And apparently those three charges produced fields that sum to zero everywhere except outside the outer surface of the shell.
It is far from obvious to me why this should be so. Inside the metal yes, but in the interior void?
@ACuriousMind if it gets reopened I will put a 500 bounty on it, so if you know the answer reopen it and post your answer :-)

3:06 PM
I'll reopen it, but I won't answer it because that'd look like bribery ;P

:-)

damn, ACM has ethics

@satan29 You need a smiley on that ...

@satan29 not sure whether that's a compliment or an insult :P

3:46 PM
Is that contagious?

@ACuriousMind It is! HTF did you find that?

@JohnRennie can you clarify your comment? I have to get to a meeting in 10mins but I actually struggled to understand the question.
I mean the question by the OP.

@ZeroTheHero I need to go. Ping me tomorrow and we can discuss it.

@JohnRennie Well, I thought about the field inside the conductor being divergence-free and curl-free by Maxwell's equation, meaning the field inside is the gradient of a harmonic function and then googled for "gradient of harmonic function zero"

3:52 PM
That's exactly the sort of insight I was hoping for :-)

@ACuriousMind Is the WZW action of the $\mathbb{R}^n$ group the Polyakov action?
Apparently it may be $U(1)^D$ instead

Well, since in the pure theory you only have the algebra-valued connection, it's not so easy to distinguish $\mathbb{R}^D$ from $\mathrm{U}(1)^D$ at the level of the action - the two look exactly the same, since the Lie algebra of $\mathrm{U}(1)^D$ is $\mathbb{R}^D$, too.

4:17 PM
True
Is the WZW model basically just Polyakov for target spaces that look like Lie groups?

4:30 PM
@JohnRennie can't do tomorrow but early next week. There is a non-zero field inside the shell (the void area). The problem is computationally very complicated and I would actually love to discuss this at greater length.
@ACuriousMind make sure you join that conversation... another meeting is stating bye!

4:52 PM
My goodness a 500 point bounty certainly rattled the cages.

user434058
5:04 PM
@JohnRennie I voted to close after I rollback-ed, since the original version was HW like enough to get closed. But then the second version was restored, and I forgot to retract my close vote.

@JohnRennie what kind of an answer are you looking for your bountied question?

user434058
@ACuriousMind That was exactly my motivation behind rolling back and closing.

because i feel like the answer given by void is valid. something along the lines of well we have a solution which happens to be E=0 inside and we know that the solution is unique so it has to be E=0
are you looking for a more intuitive physicsy answer?

5:23 PM
@JohnRennie It is an interesting problem, conceptually and computationally...

5:44 PM
@Gonenc the question was originally asked of me by a student studying for the Indian JEE exam. This is the exam Indian students take for entry to university, so while they know basic electrostatics (e.g. Gauss's law) they won't know anything beyond this. They wouldn't know the uniqueness theorem.
So I was hoping an answer could be provided using only the knowledge of electrostatics you'd expect of someone finishing school and about to start university.
There are some excellent answers that I have really enjoyed reading, but I suspect they might be a little advanced for the sort of student I have in mind.
But I have to say that's the best 500 points I have ever spent! :-)

user434058
Let $\mathbf r$ be a function of $q_1,q_2,\dots,q_n,t$, where $t$ is time. I will be referring to total time derivatives by a dot over the symbol. Now, how do I prove that$$\frac{\partial \dot{\mathbf r}}{\partial \dot{q}_j}=\frac{\partial \mathbf r}{\partial q_j}$$ where $j\in\{1,2,\dots, n\}$?

user434058
NVM, I got it.

user434058
6:20 PM
Am I right in saying that "a time independent Lagrangian necessarily implies that the Hamiltonian is a constant, however, it does not necessarily imply that the total energy is constant"?

6:56 PM
@FakeMod It depends on what you mean by "the Hamiltonian is a constant" :P
Much of classical mechanics in this area is plagued by imprecise statements. A time-independent Hamiltonian is necessarily a constant of motion, but it is not constant as a function on phase space

user434058
7:25 PM
@ACuriousMind No, I was saying that the Hamiltonian assumes a constant value over the solution path in the phase space, given the Lagrangian has no explicit time dependence. Is this true?

What is "the solution path"?
It is a constant of motion, which means it's constant along all solutions to the equations of motion

user434058
@ACuriousMind Yeah, that's what I was talking about.

user434058
@ACuriousMind The path the system takes in the phase space.

Well, there is only a unique such path after you have specified initial conditions

user434058
@ACuriousMind Yup.

user434058
7:30 PM
How do I correlate D'Alembert's principle and the principle of least action directly, without bringing in the Euler-Lagrange equations?

What do you mean by "correlating" them?

user434058
@ACuriousMind Showing that both of them are equivalent by deriving one of them from the other, and vice versa.

They are not equivalent

user434058
@ACuriousMind Aren't they equivalent at least for classical mechanics?

d'Alembert also applies to dissipative systems, while the principle of least action in its standard formulation does not because there is no potential for dissipative forces
You can derive the principle of least action from d'Alembert + Newton's laws + assumptions about the forces having potentials, but not the other way around

user434058
7:41 PM
@ACuriousMind So how do I do it, if I make the assumption that there are no dissipative forces? Also, I don't see why we'd need Newton's laws, because D'Alembert's principle is equivalent to Newton's laws (isn't it?).

user434058
@ACuriousMind Ah, I've read that before :D

user434058

user434058
7:57 PM
@ACuriousMind Hmm... so that answer uses Euler-Lagrange equations as the bridge to go from D'Alembert's principle to Lagrangian mechanics. I ask, is there any more direct way?

I don't understand what you mean by a "more direct" way

user434058
Oops, in the above comment, I meant "Action principle" rather than" Lagrangian mechanics".

user434058
Gtg

8:29 PM
hello

hi

I want to find the moment of inertia for an oblique cylinder. Any idea how to do this?

@StanShunpike Compute the integral given at the end of this Wikipedia section

@ACuriousMind ok i'll take a look. thanks!